Help with Understanding CH27 of Srednicki

[LAHLH]

Hi,

I'm reading through CH27 of Srednicki at the moment, and struggling to understand a couple of concepts.

1) He states that in the MS (bar) scheme the location of the pole in exact propagator is no longer when $$k^2=-m^2$$, where m is Lagrangian parameter usually thought of as mass. I think I understand this because we no longer are imposing the conditions $$\Pi(-m^2)=0$$ etc, as in the OS scheme, so of course there is no pole in the exact prop when $$k^2=-m^2$$. However he then goes on to say the physical mass, $$m_{ph}$$ is defined by the location of the pole: $$k^2=-m^{2}_{ph}$$. Why is the physical mass defined this way? what's so special about the place where there is a pole in exact propagator? How do these things tie in with the Lehman-Kallen form of the exact propagator that clearly shows there must be a pole when $$k^2=-m^2$$, is this 'm' in the Lehman-Kallen formula $$m_{ph}$$?

2) He then states the LSZ formula must be corrected by mutliplying it's RHS by a factor of $$\tfrac{1}{\sqrt{R}}$$ where R is the residue of the pole, and the reason he gives is that it is the field $$\tfrac{\phi(x)}{\sqrt{R}}$$ that has unit amplitude to create a one particle state. I have no idea why this is, and it would be really great if anyone could explain some more.

3) My final question is how he gets to 27.12 just by taking the log of 27.11:
He starts with, $$m^{2}_{ph}=m^2[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)]$$
Now taking logs:
$$2ln(m_{ph})=2ln(m)+ln[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)]$$

My only thought is that in the second term perhaps you could write $$1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2) =exp(\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c'))$$
since the second term is second order anyway, it kind of doesn't matter if its the real $$O(\alpha^2)$$ term form or not. Then you would have $$ln[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)]=ln( exp(\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')))=\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c'))$$

and you recover 27.12?

 

[azntoon]

Thanks so much!1) The physical mass is defined by the pole in the exact propagator because it is the mass of the particle that the exact propagator describes. In the Lehman-Kallen form, the mass is the Lagrangian parameter m, which is the same as the physical mass when the OS scheme is used. When the MS scheme is used, the Lagrangian parameter m is not necessarily equal to the physical mass, so the physical mass is found by finding the pole in the exact propagator. 2) The LSZ formula tells us that the amplitude for a process involving n incoming particles and m outgoing particles is proportional to the matrix element \langle 0| \hat{\phi}(x_1) \hat{\phi}(x_2)...\hat{\phi}(x_m)|p_1, p_2 ... p_n \rangle where the hat indicates normal ordering, and \phi(x_i) is the field operator. This matrix element is calculated using the exact propagator, and is proportional to the residue of the pole. Since the amplitude should give a unit probability of creating a one particle state, it must be rescaled by a factor of \tfrac{1}{\sqrt{R}} to account for the fact that the exact propagator gives a probability of creating a one particle state that is proportional to R, whereas the true probability should be 1. 3) The correct derivation of equation 27.12 is as follows:Start with equation 27.11:m^{2}_{ph}=m^2[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)]Now take logs on both sides:2ln(m_{ph})=2ln(m)+ln[1+\tfrac{5}{12}\alpha(ln(\mu^2/m^2)+c')+O(\alpha^2)] Now use the identity ln(1+x)=x-\tfrac{x^2}{2}+O(x^3):2ln(m_{ph})=2ln(m)+\tfrac{5